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# can anybody solve this one?

### #1

Posted 26 December 2007 - 11:25 AM

Given that when I pick a corner of A, the 3 sides connected to it read "653", find the same orientation of B, C and D as 1 string of numbers. For example, "152421563".

### #2

Posted 27 December 2007 - 02:05 AM

Usually the cubes are stacked, so that 4 faces are seen [apart from the ends]

and the idea is that each of the four sets of 4 faces show all four colors.

Are you posing the problem in such a way that all 6 faces of the four cubes

are visible, and the six sets of 4 faces show all four colors? Sounds like you

are, and that makes for an interesting puzzle. Will look at it more tonight.

Good one.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #3

Posted 27 December 2007 - 09:44 AM

Imagine that you have the layouts of 4 coloured cubes, named A, B, C and D. When the cubes are placed in (alphabetical) order in a straight line, and it is viewed from any horizontal or vertical position (front, back, top, bottom, left, right), there are always four different colours seen.

I don't understand what this means. If I put four cubes in a row and then look from the side, I can see one cube, not four. Ergo, one color, not four.

Given that when I pick a corner of A, the 3 sides connected to it read "653", find the same orientation of B, C and D as 1 string of numbers. For example, "152421563".

Not a clue as to what this is supposed to mean.

### #4

Posted 27 December 2007 - 10:53 AM

Given that when I pick a corner of A, the 3 sides connected to it read "653", find the same orientation of B, C and D as 1 string of numbers. For example, "152421563".

Not a clue as to what this is supposed to mean.

If A is oriented so that sides 653 are seen from some corner, orient B so that 152 are seen on the corresponding corner, 431 are seen on that corner of C, and 453 are seen on the corresponding corner of D. That's the notation he's asking for [but use the actual face numbers from your solution, not 152 421 and 563.]

I'm pretty sure he's asking that the top faces need to be different colors; same for the bottom, front, back, left and right faces.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #5

Posted 27 December 2007 - 11:56 AM

Clarification needed:

Usually the cubes are stacked, so that 4 faces are seen [apart from the ends]

and the idea is that each of the four sets of 4 faces show all four colors.

Are you posing the problem in such a way that all 6 faces of the four cubes

are visible, and the six sets of 4 faces show all four colors? Sounds like you

are, and that makes for an interesting puzzle. Will look at it more tonight.

Good one.

Thats correct and your deffinetly on the right track. Whoever can correctly give me the 9 correct digits in order by lets say the last day of this month, Id personally like to give them $1 USD direct to there paypal account! This is just for fun and I hope this allowed to be said. If the correct answer is stated b4 the end of this year ill more the gladly send the $1 for the work of solving it. Good luck to all and ill personally ask for your paypal email in pm to give you the $1. (Hope I havent said anything not allowed on this forum and im just trying to create more fun) Thanx peeps.

### #6

Posted 27 December 2007 - 12:06 PM

Hope I havent said anything not allowed on this forum and im just trying to create more fun

that's ok with me

rookie1ja (site admin)

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### #7

Posted 27 December 2007 - 03:33 PM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #8

Posted 28 December 2007 - 04:31 AM

### #9

Posted 28 December 2007 - 06:42 AM

Nice puzzle.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #10

Posted 05 January 2008 - 09:31 AM

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